How to solve the probability of the binomial distribution sequence below?

Question

Let $x_0$, $x_1$...$x_n$ be a sequence of independent random variables. $x_i = 1$ has probability $p$ and $x_i = 0$ has probability $1-p$. Let $k$ be the smallest integer such that $x_k = x_{k+1}$. Find probability that $x_k = 1$.

Example: $1,0,0,1,1,\cdots\quad k = 1$ and $x_k = 0$.

My idea is that the probability of $x_k = 1$ is equal to the probability that no consecutive zero occurs before, which is $1-x_k = 0$ and then this question becomes the same question.

Answer 1

Denote $r:=P(x_k=1)$ and $r_0:=P(x_k=1\mid x_0=0)$ and $r_1:=P(x_k=1\mid x_0=1)$.

Then we have the equalities:

• $r=pr_1+(1-p)r_0$
• $r_0=pr_1$
• $r_1=p+(1-p)r_0$

Do you see why, and can you take it from here?

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$\begin{array}{ccccc} & & 0\\ & 1-p\nearrow & & \nwarrow\\ \text{start } & & p\downarrow & & \uparrow1-p\\ & p\searrow & & \nearrow\\ & & 1\\ & & & p\searrow\\ & & & & x_{k}=1 \end{array}$

Answer 2

For $x_k=1$, the first $k-1$ samples must either be: an alternating sequence of $1,0$ (and $k$ is odd) or $0$ followed by an alternating sequence of $1,0$ (and $k$ is even); and of course samples $x_k, x_{k+1}$ must then be both $1$.